Problem: A big cruise ship dropped anchor off the Caribbean island of Barbuda. The heavy anchor was released from a height of $54$ meters above the water's surface, and it fell at a constant rate. After $35$ seconds, the anchor was $9$ meters below the water's surface. How fast did the anchor drop?
Solution: Let's say that the anchor dropped at a rate of $V$ meters per second. Then, it dropped $V\cdot T$ meters in $T$ seconds. In addition, we know that the anchor was released from $54$ meters above the water's surface. The anchor's elevation relative to the water's surface can be found by taking the height from which the anchor was released and subtracting the distance it dropped. We can express this with the equation $E=54-V\cdot T$, where: $E$ represents the anchor's elevation relative to the water's surface at a given time (in meters) $V$ represents the speed at which the anchor dropped (in meters per second) $T$ represents the time (in seconds) We know that after $35$ seconds $(T={35})$, the anchor was $9$ meters below the water's surface $(E={-9})$. Let's plug these values into the equation to find the value of $V$. $ \begin{aligned}{-9}&=54-V\cdot{35}\\ 35V&=63\\ V&=1.8\end{aligned}$ Therefore, the anchor dropped at a rate of $1.8$ meters per second. To find how long it took the anchor to reach the water's surface, we can plug $E=0$ into the equation and solve for $T$. $ \begin{aligned}0&=54-1.8T\\ 1.8T&=54\\ T&=30\end{aligned}$ The anchor dropped at a rate of $1.8$ meters per second. It took the anchor $30$ seconds to reach the water's surface.